Mathematics - Hyperbola Question with Solution | TestHub

On substituting $\left(\frac{a}{e},0\right)$ in $y=-2x+1$, we get $$\begin{array}{rl}0& =-\frac{2a}{e}+1\\ \Rightarrow \frac{a}{e}& =\frac{1}{2}\end{array}$$ Also, $y=-2x+1$ is tangent to hyperbola $$\begin{array}{ll}\therefore & 1=4a^2-b^2\\ \Rightarrow & \frac{1}{a^2}=4-\left(e^2-1\right)\end{array}$$
$$\begin{array}{ll}\Rightarrow & \frac{4}{e^2}=5-e^2\\ \Rightarrow & e^4-5e^2+4=0\\ \Rightarrow & \left(e^2-4\right)\left(e^2-1\right)=0\\ \Rightarrow & e=2,e=1\end{array}$$ $e=1$ gives the conic as parabola. But conic is given as hyperbola, hence $e=2$